Physics 9702 · Paper 1

Force = Rate of Change of Momentum

$F = ma$ from lesson 3.2 is a special case. The full version of Newton's second law uses momentum.
This full version is also the exam's definition of force.

The full second law

F = \dfrac{\Delta p}{\Delta t}

Symbols

$F$ = resultant force (N)
$\Delta p$ = change in momentum (kg m s⁻¹)
$\Delta t$ = time taken (s)

Newton's second law: the resultant force on an object equals the rate of change of its momentum, and they point the same way.
This is also the definition asked as “define force”: force is the rate of change of momentum (9702/22/F/M/25 Q3(b)(i)).
For a constant mass, $\Delta p = m\Delta v$ , so $F = m\dfrac{\Delta v}{\Delta t} = ma$ . That is where lesson 3.2 came from.
When the mass changes (a rocket burning fuel, rain landing in a truck), $F = ma$ no longer works and you must use $F = \Delta p / \Delta t$ .

Common mistake

On the multiple-choice paper, “force is the product of mass and acceleration” appears as an option for the definition of force. It is wrong:

ma

is a formula for the special case, not the definition (9702/13/M/J/25 Q9).

Momentum-time graphs

$F = \Delta p / \Delta t$ read as a graph: on a momentum–time graph, the gradient is the resultant force.
A straight line means a constant resultant force. A curve flattening out means the resultant force is shrinking.

On a momentum-time graph the gradient is the resultant force: F = Δp / Δt = 3.0 / 2.0 = 1.5 N.

Worked example

Gradient, then one more force

A block slides while a constant friction force of 2.0 N acts against an applied force X. The block's momentum–time graph is a straight line rising 6.0 kg m s⁻¹ every 4.0 s. Find X. (9702/21/M/J/23 Q3)

Resultant force = gradient = $\dfrac{6.0}{4.0} = 1.5 \text{ N}$ .
Resultant = applied − friction: $1.5 = X - 2.0$ , so $X = 3.5 \text{ N}$ .

Answer

X = 3.5 \text{ N}

Worked example

One change: the momentum is falling

A block moves in the direction of an applied force F while a resistive force of 5.0 N acts backwards. Its momentum falls from 2.7 to 1.5 kg m s⁻¹ in 0.40 s. Find F. (9702/13/M/J/24 Q6)

Gradient: $\dfrac{1.5 - 2.7}{0.40} = -3.0 \text{ N}$ . The resultant is 3.0 N backwards.
Taking forward as positive: $F - 5.0 = -3.0$ , so $F = 2.0 \text{ N}$ .

Answer

F = 2.0 \text{ N}

The falling graph does not mean F is negative. It means the resultant is backwards: friction is 3.0 N larger than F.

A steady stream of mass

Some questions have a continuous flow: water hitting a wall, bullets leaving a gun, air pushed by a fan.
Then $\Delta p / \Delta t$ = (mass arriving per second) × (change in velocity of that mass).

A horizontal jet of water from a pipe hitting a flat vertical wall and running down it. — A steady jet brings a certain mass of water to the wall every second, and the wall stops it. Force = rate of change of momentum: the faster the flow and the more mass arriving each second, the bigger the steady force on the wall.

Worked example

Water on a wall

Water hits a wall horizontally at 8.0 m s⁻¹ and stops. The water arrives at 5.0 kg per second. Find the force on the wall.

Each second, 5.0 kg loses 8.0 m s⁻¹ of velocity: $F = \dfrac{\Delta p}{\Delta t} = 5.0 \times 8.0 = 40 \text{ N}$ .

Answer

F = 40 \text{ N}

Your turn— tap to reveal the worked answer (9702-style)

A machine gun fires bullets of mass 0.014 kg at 640 m s⁻¹. The soldier can hold the gun against a steady force of at most 140 N. How many bullets can the gun fire each second?

One bullet: $p = 0.014 \times 640 = 8.96 \text{ kg m s}^{-1}$ .
$n \times 8.96 \le 140$ , so $n \le 15.6$ .

Answer: 15 bullets per second.

Gradient of a momentum–time graph = resultant force. Flow questions: force = mass per second × velocity change.

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Force = Rate of Change of Momentum

$F = ma$ from lesson 3.2 is a special case. The full version of Newton's second law uses momentum.
This full version is also the exam's definition of force.

The full second law

F = \dfrac{\Delta p}{\Delta t}

Symbols

$F$ = resultant force (N)
$\Delta p$ = change in momentum (kg m s⁻¹)
$\Delta t$ = time taken (s)

Newton's second law: the resultant force on an object equals the rate of change of its momentum, and they point the same way.
This is also the definition asked as “define force”: force is the rate of change of momentum (9702/22/F/M/25 Q3(b)(i)).
For a constant mass, $\Delta p = m\Delta v$ , so $F = m\dfrac{\Delta v}{\Delta t} = ma$ . That is where lesson 3.2 came from.
When the mass changes (a rocket burning fuel, rain landing in a truck), $F = ma$ no longer works and you must use $F = \Delta p / \Delta t$ .

Common mistake

On the multiple-choice paper, “force is the product of mass and acceleration” appears as an option for the definition of force. It is wrong:

ma

is a formula for the special case, not the definition (9702/13/M/J/25 Q9).

Momentum-time graphs

$F = \Delta p / \Delta t$ read as a graph: on a momentum–time graph, the gradient is the resultant force.
A straight line means a constant resultant force. A curve flattening out means the resultant force is shrinking.

On a momentum-time graph the gradient is the resultant force: F = Δp / Δt = 3.0 / 2.0 = 1.5 N.

Worked example

Gradient, then one more force

Resultant force = gradient = $\dfrac{6.0}{4.0} = 1.5 \text{ N}$ .
Resultant = applied − friction: $1.5 = X - 2.0$ , so $X = 3.5 \text{ N}$ .

Answer

X = 3.5 \text{ N}

Worked example

One change: the momentum is falling

A block moves in the direction of an applied force F while a resistive force of 5.0 N acts backwards. Its momentum falls from 2.7 to 1.5 kg m s⁻¹ in 0.40 s. Find F. (9702/13/M/J/24 Q6)

Gradient: $\dfrac{1.5 - 2.7}{0.40} = -3.0 \text{ N}$ . The resultant is 3.0 N backwards.
Taking forward as positive: $F - 5.0 = -3.0$ , so $F = 2.0 \text{ N}$ .

Answer

F = 2.0 \text{ N}

The falling graph does not mean F is negative. It means the resultant is backwards: friction is 3.0 N larger than F.

A steady stream of mass

Some questions have a continuous flow: water hitting a wall, bullets leaving a gun, air pushed by a fan.
Then $\Delta p / \Delta t$ = (mass arriving per second) × (change in velocity of that mass).

Worked example

Water on a wall

Water hits a wall horizontally at 8.0 m s⁻¹ and stops. The water arrives at 5.0 kg per second. Find the force on the wall.

Each second, 5.0 kg loses 8.0 m s⁻¹ of velocity: $F = \dfrac{\Delta p}{\Delta t} = 5.0 \times 8.0 = 40 \text{ N}$ .

Answer

F = 40 \text{ N}

Your turn— tap to reveal the worked answer (9702-style)

A machine gun fires bullets of mass 0.014 kg at 640 m s⁻¹. The soldier can hold the gun against a steady force of at most 140 N. How many bullets can the gun fire each second?

One bullet: $p = 0.014 \times 640 = 8.96 \text{ kg m s}^{-1}$ .
$n \times 8.96 \le 140$ , so $n \le 15.6$ .

Answer: 15 bullets per second.

Gradient of a momentum–time graph = resultant force. Flow questions: force = mass per second × velocity change.